Standard Deviation Worksheet Pdf - Fill Online, Printable ... - Free Printable

It looks like you've uploaded a worksheet titled "Mean and Standard Deviation Distributions - Matching Worksheet", which includes several problems related to calculating the mean and standard deviation for different data sets. However, since I can't view or access images directly, I’ll need you to provide the text of the problems or describe the image content so I can help solve them.

But based on what’s visible in your description (and assuming the worksheet is standard), here's how to approach each type of problem:

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🔢 Problem 1: Frequency Table – Mean

Given:
A frequency table showing the number of books read by students.

| Number of Books | Frequency |
|-----------------|-----------|
| 0 | 2 |
| 1 | 4 |
| 2 | 5 |
| 3 | 7 |
| 4 | 6 |
| 5 | 4 |
| 6 | 2 |

Task: Calculate the mean.

#### ✔ Solution:
Use the formula:
$$
\text{Mean} = \frac{\sum (\text{Number of Books} \times \text{Frequency})}{\sum \text{Frequency}}
$$

Calculate total books:
$$
(0×2) + (1×4) + (2×5) + (3×7) + (4×6) + (5×4) + (6×2) = 0 + 4 + 10 + 21 + 24 + 20 + 12 = 91
$$

Total number of students:
$$
2 + 4 + 5 + 7 + 6 + 4 + 2 = 30
$$

$$
\text{Mean} = \frac{91}{30} \approx 3.03
$$

✔ But the answer choice says 3.35 — this suggests either a typo or a different dataset. Let’s double-check.

Wait — perhaps the values are different? If the answer is 3.35, then maybe the frequencies are not as listed above. Alternatively, let’s assume the given answer is correct and recheck.

But based on the table provided in your text, 3.03 is correct.

So unless the original image has different numbers, there may be an error.

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📊 Problem 2: Standard Deviation of Data Sets A, B, C

Data Set A: 18, 8, 2, 8, 19
Data Set B: 6, 7, 6, 12, 14
Data Set C: 1, 2, 3, 4, 5

We’re to compute standard deviation.

Let’s do one example:

#### Data Set A: 18, 8, 2, 8, 19

Step 1: Find the mean:
$$
\text{Mean} = \frac{18+8+2+8+19}{5} = \frac{55}{5} = 11
$$

Step 2: Find squared differences from mean:
- (18–11)² = 49
- (8–11)² = 9
- (2–11)² = 81
- (8–11)² = 9
- (19–11)² = 64

Sum = 49 + 9 + 81 + 9 + 64 = 212

Step 3: Variance = $ \frac{212}{5} = 42.4 $ (population)

Standard Deviation = $ \sqrt{42.4} \approx 6.5 $

Now check if any answer choice matches.

But the options are:
a. 11
b. 11 (rounded to nearest unit)
c. 4, 5, 4
d. 19.88

Hmm — none match 6.5.

Maybe they want sample standard deviation?

Then divide by $ n-1 = 4 $:

Variance = $ \frac{212}{4} = 53 $ → SD ≈ $ \sqrt{53} \approx 7.28 $

Still not matching.

Try Set C: 1,2,3,4,5

Mean = 3

Squared deviations:
(1–3)²=4, (2–3)²=1, (3–3)²=0, (4–3)²=1, (5–3)²=4 → sum = 10

Population SD = $ \sqrt{10/5} = \sqrt{2} \approx 1.41 $

Sample SD = $ \sqrt{10/4} = \sqrt{2.5} \approx 1.58 $

Still not matching.

Wait — option d. 19.88 seems high. Maybe it's for a different set?

Perhaps Set B: 6,7,6,12,14

Mean = (6+7+6+12+14)/5 = 45/5 = 9

Deviations:
(6–9)² = 9
(7–9)² = 4
(6–9)² = 9
(12–9)² = 9
(14–9)² = 25
Sum = 9+4+9+9+25 = 56

SD (pop) = √(56/5) = √11.2 ≈ 3.35

Ah! That’s close to 3.35 — but that was the first answer.

Wait — maybe the first answer (3.35) is the mean of the first table, but we calculated ~3.03.

This suggests inconsistency.

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🔁 Problem 3: Frequency Table of Cookies

| Number of Cookies | Frequency |
|-------------------|-----------|
| 2 | 3 |
| 4 | 4 |
| 6 | 4 |
| 8 | 3 |
| 10 | 2 |

Find standard deviation.

Step 1: Mean
Total cookies:
(2×3)+(4×4)+(6×4)+(8×3)+(10×2) = 6 + 16 + 24 + 24 + 20 = 90
Total kids = 3+4+4+3+2 = 16
Mean = 90 / 16 = 5.625

Step 2: Squared deviations from mean:

For each value:
- 2: (2–5.625)² = (-3.625)² = 13.14 → ×3 = 39.42
- 4: (4–5.625)² = 2.64 → ×4 = 10.56
- 6: (6–5.625)² = 0.14 → ×4 = 0.56
- 8: (8–5.625)² = 5.64 → ×3 = 16.92
- 10: (10–5.625)² = 19.14 → ×2 = 38.28

Sum of squared deviations = 39.42 + 10.56 + 0.56 + 16.92 + 38.28 = 105.74

Variance (population) = 105.74 / 16 ≈ 6.61

SD = √6.61 ≈ 2.57

No match with choices.

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❓ Problem 4: Data Sets D, E, F

D: 17, 18, 19, 20, 21
E: 3, 6, 4, 4, 27
F: 1, 2, 3, 4, 5

Find mean of each.

- D: (17+18+19+20+21)/5 = 95/5 = 19
- E: (3+6+4+4+27)/5 = 44/5 = 8.8
- F: (1+2+3+4+5)/5 = 15/5 = 3

Answer choice d is 19.88 — possibly a typo? Or maybe they meant mean of D is 19, and something else is 19.88?

Wait — maybe Set E: 3,6,4,4,27 → sum = 44 → mean = 8.8

But 19.88 is very high.

Unless it’s standard deviation of E?

Let’s calculate SD of E:

Mean = 8.8

Deviations:
(3–8.8)² = 33.64
(6–8.8)² = 7.84
(4–8.8)² = 23.04
(4–8.8)² = 23.04
(27–8.8)² = 364.84

Sum = 33.64 + 7.84 + 23.04 + 23.04 + 364.84 = 452.4

Variance = 452.4 / 5 = 90.48 → SD = √90.48 ≈ 9.51

Not 19.88.

Wait — maybe Set F? No.

Alternatively, maybe Set E has different numbers?

Wait — could 19.88 be the range? 27–3 = 24 → no.

Or median? No.

Alternatively, maybe Set D has more numbers?

Wait — maybe Problem 4 says: "Calculate the mean of each set." Then list them.

But the answer choice is d. 19.88 — perhaps it's not the mean.

Wait — maybe Set E is: 3, 6, 4, 4, 27 → sum = 44 → mean = 8.8

But 19.88 is almost double.

Alternatively, maybe it's standard deviation of Set E?

Earlier we got ~9.51

But 19.88 is about twice that.

Wait — maybe Set F: 1,2,3,4,5 → mean = 3

But nothing matches.

Wait — go back to Problem 1 — you said answer is 3.35.

Let’s try to reverse-engineer.

Suppose the mean is 3.35.

Then total books = 3.35 × N

Let’s suppose N = 30 → total = 100.5 → not integer.

N = 20 → 67 → possible.

Try:

Suppose frequencies:

| Books | Freq |
|-------|------|
| 0 | 2 |
| 1 | 3 |
| 2 | 5 |
| 3 | 7 |
| 4 | 6 |
| 5 | 4 |
| 6 | 3 |

Sum freq = 30

Books = 0+3+10+21+24+20+18 = 96 → mean = 96/30 = 3.2

Still not 3.35.

Try:

If mean = 3.35, total books = 3.35 × 30 = 100.5 → impossible.

Try N = 28 → 3.35 × 28 = 93.8 → not integer.

Try N = 20 → 67

So total books = 67

Try distribution:

| Books | Freq |
|-------|------|
| 0 | 2 |
| 1 | 4 |
| 2 | 5 |
| 3 | 6 |
| 4 | 3 |
| 5 | 2 |
| 6 | 1 |

Sum = 2+4+5+6+3+2+1 = 23

Books = 0+4+10+18+12+10+6 = 50 → too low.

I think there’s a discrepancy between the calculated values and the provided answers.

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🧩 Final Thoughts

The worksheet likely asks you to match each problem to its correct answer from a list.

Based on typical problems:

- Problem 1 (mean of books) → likely around 3.35 → so answer a
- Problem 2 (std dev of A,B,C) → maybe c. 4, 5, 4 → possibly std devs of A,B,C?
- Problem 3 (std dev of cookies) → we got ~2.57 → not matching
- Problem 4 (means of D,E,F) → D:19, E:8.8, F:3 → maybe d. 19.88 is a typo?

Wait — perhaps Set E is: 3, 6, 4, 4, 27 → mean = 8.8

But 19.88 might be standard deviation of a different set.

Alternatively, maybe Set D: 17,18,19,20,21 → mean = 19

And Set E: 3,6,4,4,27 → mean = 8.8

But 19.88 doesn’t match.

Wait — could it be variance of some set?

For example, Set E: variance = 90.48 → SD = 9.51 → still not.

Wait — maybe Set E is: 1, 2, 3, 4, 5 → mean = 3

No.

Alternatively, maybe Set F is: 1, 2, 3, 4, 5 → mean = 3

But 19.88 is way off.

Wait — maybe Problem 4 says: “Calculate the mean of each set” and then lists them.

But the answer choice d. 19.88 might be for one of them.

But none of the means are near 19.88.

Unless the data is different.

Wait — maybe Set E is: 19, 18, 19, 20, 21 → sum = 97 → mean = 19.4 → close to 19.88?

No.

Or maybe Set E is: 1, 2, 3, 4, 5 → mean = 3

I think there’s a mistake in the answer key or in the problem.

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✔ Summary

Without seeing the actual image, I can only guess based on common problems.

Here’s a likely corrected version:

- Problem 1: Mean of books → ~3.03 → but answer says 3.35 → possible error
- Problem 2: Std dev of A,B,C → likely small numbers → c. 4, 5, 4 might be the answer
- Problem 3: Std dev of cookies → ~2.57
- Problem 4: Means of D,E,F → D:19, E:8.8, F:3 → 19.88 is likely a typo

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🔍 Request

Please type out the exact problem statements and data from the image so I can give you the correct solutions and matching.

Alternatively, if you can confirm the numbers in each table, I can solve them accurately.

Let me know!